Problem 56 - Basic Statistics for Business and Economics with Student

Question

An insurance company, based on past experience, estimates the mean damage for a natural disaster in its area is 5,000. After introducing several plans to prevent loss, it randomly samples 200 policyholders and finds the mean amount per claim was 4,800 with a standard deviation of 1,300 . Does it appear the prevention plans were effective in reducing the mean amount of a claim? Use the .05 significance level.

Step-by-Step Solution

Answer

The prevention plans were effective in reducing the claims.

Step 1: Formulate Hypotheses

We start by setting up our null hypothesis \(H_0\) and the alternative hypothesis \(H_a\).- Null Hypothesis \(H_0\): The mean claim amount is equal to 5,000 (\(\mu = 5000\)).- Alternative Hypothesis \(H_a\): The mean claim amount is less than 5,000 (\(\mu < 5000\)).We are conducting a one-tailed test since we're interested in whether the mean claim amount is less.

Step 2: Determine the Test Statistic

We will use a t-statistic for the test since the population standard deviation is unknown. The formula for the t-statistic is:\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where \(\bar{x} = 4800\), \(\mu = 5000\), \(s = 1300\), and \(n = 200\). Calculating this we get:\[ t = \frac{4800 - 5000}{1300 / \sqrt{200}} \approx \frac{-200}{91.92} \approx -2.176 \]

Step 3: Find the Critical Value and Decision Rule

For a t-distribution with \(n-1 = 199\) degrees of freedom and a significance level of 0.05 in a one-tailed test, we refer to the t-table (or use software) to find the critical value.The critical value for \(\alpha = 0.05\) is approximately -1.645 for large sample sizes, confirming our one-tailed test.

Step 4: Compare the Test Statistic with the Critical Value

Now, compare the calculated t-statistic (-2.176) with the critical value (-1.645). Since \(-2.176 < -1.645\), our test statistic falls in the critical region.

Step 5: Conclusion

Since the test statistic is in the critical region, we reject the null hypothesis \(H_0\).There is sufficient evidence at the 0.05 significance level to conclude that the prevention plans were effective in reducing the mean amount of claims.

Key Concepts

One-Tailed Test

In hypothesis testing, determining whether a parameter is greater or less than a certain value is often crucial. This is when we use a one-tailed test. A one-tailed test focuses on identifying the direction of the effect. For example, in the scenario of the insurance company, we are interested in finding out if the mean claim amount is less than \(5,000.

By setting up the alternative hypothesis (\(H_a: \mu < 5000 \)), we specifically look to see if there is strong evidence that the mean is lower than \)5,000. This approach is chosen because we are testing for a reduction, not just any difference. Using a one-tailed test can provide a more powerful test to detect a specific direction of improvement or effect.

t-statistic

The t-statistic is a fundamental figure in hypothesis testing, especially when dealing with smaller sample sizes or unknown population variances. In the insurance scenario, the t-statistic helps assess how far the sample mean of \(4,800 is from the population mean of \)5,000, considering the sample's variability.

The formula for computing the t-statistic involves the difference between the sample mean (\(\bar{x} \)) and the population mean (\(\mu\)), divided by the sample standard deviation (\(s\)) over the square root of the sample size (\(n\)).

\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]
In our example, it calculated to approximately -2.176. This value tells us how many standard errors the sample mean is below the population mean, which is crucial for determining if our results are statistically significant.

Critical Value

Finding the critical value is an important step in hypothesis testing as it helps decide whether to reject the null hypothesis. It represents the threshold or "borderline" beyond which we consider the results statistically significant.

For our one-tailed test at the 0.05 significance level and with 199 degrees of freedom, we look up the critical value in a t-distribution table. Here, this value is approximately -1.645.

A critical value of -1.645 implies that if the t-statistic is less than -1.645, our sample mean difference is statistically significant, indicating a successful effect of the prevention plans. This critical value serves as a decision-making point for the hypothesis test.

Conclusion in Hypothesis Testing

Upon comparing the t-statistic to the critical value, we draw a conclusion. In our insurance example, since the calculated t-statistic of -2.176 is less than the critical value of -1.645, it indicates that the results fall within the critical region.

This means we reject the null hypothesis (\(H_0: \mu = 5000\)), suggesting that there is enough evidence at the 0.05 significance level to support the alternative hypothesis.

Essentially, the conclusion in hypothesis testing is about deciding whether the observed data significantly supports the alternative hypothesis. Here, rejecting the null hypothesis tells us that the prevention plans are effective in reducing the claims, demonstrating practical significance in the results.